Solução_Calculo_Stewart_6e

F.
CHAPTER 5 REVIEW ¤ 259
57. Note that r(t) =b 0 (t),whereb(t) =the number of barrels of oil consumed up to time t. So, by the Net Change Theorem,
8
0
r(t) dt = b(8) − b(0) represents the number of barrels of oil consumed from Jan. 1, 2000, through Jan. 1, 2008.
59. We use the Midpoint Rule with n =6and ∆t = 24 − 0
6
=4. The increase in the bee population was
TX.10
24
0 f −1 (y) dy = 2 . 0 3
r(t) dt ≈ M6 =4[r(2) + r(6) + r(10) + r(14) + r(18) + r(22)]
≈ 4[50 + 1000 + 7000 + 8550 + 1350 + 150] = 4(18,100) = 72,400
61. Let u =2sinθ. Thendu =2cosθdθand when θ =0, u =0;whenθ = π 2 , u =2.Thus,
π/2
f(2 sin θ)cosθdθ= 2
f(u) 1
du
= 1 2 f(u) du = 1 2 f(x) dx = 1 (6) = 3.
0 0 2 2 0 2 0 2
63. Area under the curve y =sinhcx between x =0and x =1is equal to 1 ⇒
1 sinh cx dx =1 ⇒ 1 1
0 c cosh cx =1 ⇒ 1 (cosh c − 1) = 1 ⇒
0 c
cosh c − 1=c ⇒ cosh c = c +1. From the graph, we get c =0and
c ≈ 1.6161,butc =0isn’t a solution for this problem since the curve
y =sinhcx becomes y =0and the area under it is 0. Thus,c ≈ 1.6161.
65. Using FTC1, we differentiate both sides of the given equation, x
0 f(t) dt = xe2x + x
0 e−t f(t) dt, and get
f(x) =e 2x +2xe 2x + e −x f(x) ⇒ f(x) 1 − e −x = e 2x +2xe 2x ⇒ f(x) = e2x (1 + 2x)
1 − e −x .
67. Let u = f(x) and du = f 0 (x) dx. So2 b
a f(x)f 0 (x) dx =2 f(b)
f(a) udu= u 2 f(b)
f(a) =[f(b)]2 − [f(a)] 2 .
69. Let u =1− x. Thendu = −dx,so 1
0 f(1 − x) dx = 0
1 f(u)(−du) = 1
71. The shaded region has area 1
0 f(x) dx = 1 3 . The integral 1
0 f −1 (y) dy
gives the area of the unshaded region, which we know to be 1 − 1 3 = 2 3 .
So 1
f(u) du = 1
f(x) dx.
0 0